Answer :

Itz5151

Answer:

5g

Explanation:

Law of conservation of mass states that the mass of reactants and products are constant in a reaction.

  • Total Mass(Reactants) = Mass(Products)

Let x be the mass of ethane

x+17.11 =15.69 +6.42 = 22.11

x = 5 g

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There are Two methods:

First one = [tex]\sf{According~to~law~of ~ conversation }[/tex]

Law of conservation of mass states that the mass of reactants and products are constant or equal in a reaction.

  • Therefore, Total Mass of Reactant = Total Mass of Products

  • Let x be the mass of ethane.

Chemical equation:-

  • [tex]\bf{C_{2}H_{5} }[/tex] + [tex]\bf{O_{2} → }[/tex] [tex]\bf{CO_{2} }[/tex] + [tex]\bf{H_{2}O }[/tex]

  • x + 17.11g = 15.69g + 6.42g

  • x = 22.11g - 17.11g

  • x = 5.00 gram.

  • Therefore, 5.00g of ethane burns in this reaction.

Second method : According to mole concept:-

To find the mass of ethane [tex]\bf{C_{2}H_{4} }[/tex] that burns in the reaction, we need to calculate the mass of ethane consumed based on the mass of the products formed.

From the balanced chemical equation:

1 mole of [tex]\bf{C_{2}H_{4} }[/tex] reacts with 3 moles of O2 to produce 2 moles of CO2 and 2 moles of H2O.

So the molar masses are:

  • [tex]\bf{C_{2}H_{5} }[/tex] : 2×12.01 + 4×1.01 = 28.05 g/mol
  • [tex]\bf{O_{2} }[/tex]: 2×16.00 = 32.00 g/mol
  • [tex]\bf{CO_{2} }[/tex]: 12.01 + 2×16.00 = 44.01 g/mol
  • [tex]\bf{H_{2}O }[/tex]: 2×1.01 + 16.00 = 18.02 g/mol

Using the given masses:

  • Initial mass of ethane = ?g
  • Initial mass of oxygen = 17.11g

Final masses of products:

  • CO2 = 15.69g
  • H2O = 6.42g

Now, let's calculate the moles of oxygen consumed, which is the limiting reactant:

  • Moles of [tex]\bf{O_{2} }[/tex] = Mass of [tex]\sf{O_{2} }[/tex] / Molar mass of [tex]\sf{O_{2} }[/tex] [tex]\sf{\frac{17.11g}{32.00 g/mol}~ }[/tex] ≈ 0.535 mol

According to the balanced equation, 1 mole of ethane reacts with 3 moles of oxygen.

So, the moles of ethane consumed would be:

  • Moles of [tex]\bf{C_{2}H_{4} }[/tex] consumed = [tex]\sf{\frac{0.535~mol~O_{2}}{3~ mol ~O_{2}/mol }~ }[/tex] ≈ 0.178 mol

Now, let's convert moles of ethane to grams:

  • Mass of [tex]\bf{C_{2}H_{4} }[/tex] consumed = Moles of [tex]\sf{C_{2}H_{4} }[/tex] consumed * Molar mass of C2H4

  • Mass of [tex]\bf{C_{2}H_{4} }[/tex] consumed ≈ (0.178 mol) × (28.05 g/mol) ≈ 5.00g

Therefore, approximately 5.00g of ethane burns in this reaction.

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